What is the derivative of $f(x) = arcsin(2x^3 - 1)$ ?

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Answer 1 · 2018-04-13T08:43:57

$(df)/(dx)=(3x^2)/(xsqrt(x(1-x^3))$

We can use here chain formula. As differential of $arcsinx$ is $1/sqrt(1-x^2)$

derivative of $f(x)=arcsin(2x^3-1)$ is

$(df)/(dx)=1/sqrt(1-(2x^3-1)^2)*d/(dx)(2x^3-1)$

= $1/sqrt(1-4x^6+4x^3-1)*6x^2$

= $(3x^2)/(xsqrt(x(1-x^3))$

Answer 2 · 2018-04-13T12:32:13

$(3x^2)/sqrt(x^3(1-x^3)$

We use the chain rule, which states that,

$dy/dx=dy/(du)*(du)/dx$

Let $u=2x^3-1,:.(du)/dx=6x^2$ .

Then $y=arcsinu,=>dy/(du)=1/sqrt(1-u^2)$ .

Combining together, we get,

$dy/dx=1/sqrt(1-u^2)*6x^2$

$=(6x^2)/sqrt(1-u^2)$

Final step is to substitute back $u=2x^3-1$ , and we get,

$=(6x^2)/sqrt(1-(2x^3-1)^2)$

$=(3x^2)/sqrt(x^3(1-x^3)$

What is the derivative of f(x) = arcsin(2x^3 - 1)f(x) = arcsin(2x^3 - 1)?